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Question 1 of 3
Greg earns $20$20 per hour for the first 33 hours and $15$15 per hour after that. How many hours should he work to earn $135$135?
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Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.
Keywords for Word Problems |
Keyword
|
Meaning
|
Certain number |
Unknown variable (x,y,z,x,y,z, etc) |
is the same as, The answer is, is equal |
== |
sum, and, plus, more than |
++ |
subtracted, less, minus |
-− |
times, product, multiplied, twice |
×× |
half of, divide by |
÷÷ |
First, label the values and form an equation using the information given.
Extra hours needed to earn $135=h$135=h
$20$20 per hour for 33 hours |
and |
$15$15 per hour for the extra hours |
should make |
$135$135 |
3(20)3(20) |
++ |
15h15h |
== |
135135 |
3(20)+15h3(20)+15h |
== |
135135 |
60+15h60+15h |
== |
135135 |
To solve for hh, it needs to be alone on one side.
Start by moving 6060 to the other side by subtracting 6060 from both sides of the equation.
60+1560+15hh |
== |
135135 |
60+1560+15h -60 |
= |
135 -60 |
15h |
= |
75 |
60-60 cancels out |
Next, remove 15 by dividing both sides of the equation by 15.
15h |
= |
75 |
15h÷15 |
= |
75÷15 |
h |
= |
5 |
15÷15 cancels out |
Add h=5 to the first 3 hours.
Total hours needed |
= |
3+h |
|
= |
3+5 |
|
= |
5 |
Greg needs to work 8 hours to earn $135.
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Question 2 of 3
A set of 4 new tires and a $90-wheel alignment costs $898 in total. How much does each tire cost?
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Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.
Keywords for Word Problems |
Keyword
|
Meaning
|
Certain number |
Unknown variable (x,y,z, etc) |
is the same as, The answer is, is equal |
= |
sum, and, plus, more than |
+ |
subtracted, less, minus |
- |
times, product, multiplied, twice |
× |
half of, divide by |
÷ |
First, label the values and form an equation using the information given.
4 tires |
and |
a $90-wheel alignment |
costs |
$898 |
4t |
+ |
90 |
= |
898 |
To solve for t, it needs to be alone on one side.
Start by moving 90 to the other side by subtracting 90 from both sides of the equation.
4t +90 |
= |
898 |
4t +90 -90 |
= |
898 -90 |
4t |
= |
808 |
90-90 cancels out |
Next, remove 4 by dividing both sides of the equation by 4.
4t |
= |
808 |
4t÷4 |
= |
808÷4 |
t |
= |
$202 |
4÷4 cancels out |
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Question 3 of 3
A hire car costs $95 per day plus 80¢ per kilometre it is driven. If the total cost of hiring the car was $175, how many kilometres was it driven?
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Inverse Operations
When moving a term to the other side of an equation, the operation is inversed.
Keywords for Word Problems |
Keyword
|
Meaning
|
Certain number |
Unknown variable (x,y,z, etc) |
is the same as, The answer is, is equal |
= |
sum, and, plus, more than |
+ |
subtracted, less, minus |
- |
times, product, multiplied, twice |
× |
half of, divide by |
÷ |
First, make sure all units are the same. Convert the cents to dollars knowing that $1=100¢.
Cost of each kilometre the car is driven |
= |
80¢ |
|
= |
80¢÷100¢ |
|
= |
$0.8 |
Now, label the values and form an equation using the information given.
Number of kilometres driven=x
A hire car costs $95 per day |
plus |
80¢ per kilometre. |
The total cost of hiring the car was |
$175 |
95 |
+ |
0.8x |
= |
175 |
To solve for x, it needs to be alone on one side.
Start by moving 95 to the other side by subtracting 95 from both sides of the equation.
95+0.8x |
= |
175 |
95+0.8x -95 |
= |
175 -95 |
0.8x |
= |
80 |
95-95 cancels out |
Next, remove 0.8 by dividing both sides of the equation by 0.8.
0.8x |
= |
80 |
0.8x÷0.8 |
= |
80÷0.8 |
x |
= |
100 |
0.8÷0.8 cancels out |
The hire car was driven for 100 kilometres.